如何将 bash 命令的输出作为参数传递给另一个？
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how do I pass the output of a bash command as parameters to another?
提问by AriehGlazer
I have a program that returns me a list of file names. I need to pass these file's contents to another program.
我有一个程序可以返回文件名列表。我需要将这些文件的内容传递给另一个程序。
I know how to use catto pass content:
我知道如何使用cat来传递内容：
cat file1 file2 file3 | some_program
what I want to do is something like:
我想做的是：
list=`get_file_list`
cat ${list} | some_program
Though doing the above will simply pass the content of the list variable, rather than the files' content.
尽管执行上述操作只会传递列表变量的内容，而不是文件的内容。
回答by JayZee
To pass the output of,a program to another you can use xargs, for example
要将一个程序的输出传递给另一个程序，您可以使用 xargs，例如
find . -name "myfile*" -print | xargs grep "myword"
This one searches for files called myfile* and look for a key inside of them.
这将搜索名为 myfile* 的文件并在其中查找密钥。
